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(i) What is the length of the pendulum of a clock which beats seconds, given that the acceleration due to gravity at a place is equal to 9.8ms^(-2). (ii) Find the time period of another pendulum of length equal to half of it at the same place. (iii) A second pendulum is taken into a mine of depth 640 m and oscillated. What is its time period? (iv) The length of a seconds pendulum at a place is 1.02 m. Find the time period of another pendulum of length0.51 m at the same place. |
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Answer» Solution :The period of simple pendulum is given by `T=2pisqrt(l/g)implies` The length of the pendulum `=1=(gT^(2))/(4pi^(2))implies1=(9.8xx2^(2))/(4xx(3.14)^(2))` `l=(9.8xx2^(2))/(4xx9.86)=0.994m~=1m` (ii) The period of a simple pendulum is given by `T=2pisqrt(l/g)`........1 `"":.g=4pi^(2)l/(T^(2))` i.e.`l/(T^(2))=g/((4p)^(2))=` constant at a given place (2) `(l_(1))/(T_(1)^(2))=(l_(2))/(T_(2)^(2))implies1.0/(2^(2))=0.50/(T_(2)^(2))`or `T_(2)^(2)=0.50/1.01xx4=2` `:.T_(2)=sqrt(2)=1.414s` (iii) The TIME period of a simple pendulum is `T=2pisqrt(l/g)` When a SECONDS pendulum is taken into a mine g changes (l constant) The ratio of its time periods is `(T_(1))/(T_(2))=sqrt((g_(2))/(g_(1)))` On the SURFACE of the earth `g_(1)=g` and in the mine `g_(2)=g(1-d/R)=(T_(1))/(T_(2))=sqrt((g(1-d/R))/g)"or"(T_(1))/(T_(2))=sqrt(1-d/R)` Here `T_(1)=2s,d=640m=0.640km,` `R=` Radius of the earth `=6400km,T_(2)=?` `2/(T_(2))=sqrt(1-0.640/6400)=sqrt(1-0.0001)` `:.T_(2)=2/(sqrt(1-0.0001))=2.0001s` (iv) The perido of a simple pendulum is given by `T=2pisqrt(l/g)`..........1 i.e `l/(T^(2))=g/(4pi^(2))=` constant at a give place .....2 If `T_(1)` and `T_(2)` are the time periods of the pendulum for length `l_(1)` and `l_(2)` From eq. (2) we can write `(l_(1))/(T_(1)^(2))=(l_(2))/(T_(2)^(2))`.....3 Here `T_(1)=2s`, when `l_(1)=1.02m` when `l_(2)=0.51m,T_(2)=?` Using eq. 3 `1.02/(2^(2))=0.51/(T_(2)^(2))` or `T_(2)^(2)=0.51/1.02xx4=2` `:.T_(2)=sqrt(2)=1.414s`. When the length increasses the time period of the pendulum increases. So the clock LOSES time or RUNS slow. |
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